Double Integrity!

Calculus Level 2

∫ y = 1 2 ∫ x = 0 3 ( 1 + 8 x y ) d x d y = ? \huge \int _{ y=1 }^{ 2 } \int _{ x=0 }^{ 3 } (1+8xy) \, dx \, dy = \, ?


The answer is 57.

Swapnil Das by Swapnil Das , 20, India

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2 solutions

Chew-Seong Cheong
Mar 18, 2016

∫ 1 2 ∫ 0 3 ( 1 + 8 x y ) d x d y = ∫ 1 2 [ x + 4 x 2 y ] 0 3 d y = ∫ 1 2 ( 3 + 36 y ) d y = [ 3 y + 18 y 2 ] 1 2 = 6 + 72 − 3 − 18 = 57 \begin{aligned} \int_1^2 \int_0^3 (1+8xy)dxdy & = \int_1^2 \left[x+4x^2y\right]_0^3 dy \\ & = \int_1^2 \left(3+36y\right) dy \\ & = \left[3y + 18y^2\right]_1^2 \\ & = 6 + 72 - 3 - 18 \\ & = \boxed{57} \end{aligned}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Simple standard approach.

Aareyan Manzoor
Mar 18, 2016

∫ 1 2 ∫ 0 3 ( 1 + 8 x y ) d x d y = ∫ 1 2 ∫ 0 3 1 d x d y + ∫ 1 2 ∫ 0 3 8 x y d x d y = ∫ 1 2 ∫ 0 3 d x d y + 8 ∫ 1 2 ∫ 0 3 x y d x d y = ∫ 1 2 d x ∫ 0 3 d y + 8 ∫ 1 2 x d x ∫ 0 3 y d y = ( 1 ) ( 3 ) + 8 ( 3 2 ) ( 9 2 ) = 3 + 54 = 57 \int_1^2\int_0^3 (1+8xy)dxdy\\=\int_1^2\int_0^3 1 dxdy+\int_1^2\int_0^3 8xy dxdy\\= \int_1^2\int_0^3 dxdy+8\int_1^2\int_0^3xy dxdy\\=\int_1^2 dx\int_0^3dy+8\int_1^2 xdx\int_0^3ydy\\=(1)(3)+8\left(\dfrac{3}{2}\right)\left(\dfrac{9}{2}\right)\\=3+54\\=\boxed{57}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Simple standard approach.

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